x^2+40x-351=0

Solution for x^2+40x-351=0 equation:

x^2+40x-351=0

a = 1; b = 40; c = -351;
Δ = b2-4ac
Δ = 402-4·1·(-351)
Δ = 3004
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3004}=\sqrt{4*751}=\sqrt{4}*\sqrt{751}=2\sqrt{751}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{751}}{2*1}=\frac{-40-2\sqrt{751}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{751}}{2*1}=\frac{-40+2\sqrt{751}}{2}
$